Problem: Simplify and expand the following expression: $ \dfrac{2r - 5}{r - 1}-\dfrac{r - 4}{r - 9} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(r - 1)(r - 9)$ Multiply the first term by $\dfrac{r - 9}{r - 9}$ $ \begin{align*} \dfrac{2r - 5}{r - 1} \times \dfrac{r - 9}{r - 9} & = \dfrac{(2r - 5)(r - 9)}{(r - 1)(r - 9)} \\ & = \dfrac{2r^2 - 23r + 45}{(r - 1)(r - 9)}\end{align*} $ Multiply the second term by $\dfrac{r - 1}{r - 1}$ $ \begin{align*} \dfrac{r - 4}{r - 9} \times \dfrac{r - 1}{r - 1} & = \dfrac{(r - 4)(r - 1)}{(r - 9)(r - 1)} \\ & = \dfrac{r^2 - 5r + 4}{(r - 9)(r - 1)}\end{align*} $ Now we have: $ = \dfrac{2r^2 - 23r + 45}{(r - 1)(r - 9)} - \dfrac{r^2 - 5r + 4}{(r - 9)(r - 1)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{2r^2 - 23r + 45 - (r^2 - 5r + 4)}{(r - 1)(r - 9)} $ $ = \dfrac{2r^2 - 23r + 45 - r^2 + 5r - 4}{(r - 1)(r - 9)} $ $ = \dfrac{r^2 - 18r + 41}{(r - 1)(r - 9)}$ Expand the denominator: $ = \dfrac{r^2 - 18r + 41}{r^2 - 10r + 9}$